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Q.

A uniform solid sphere of mass m and radius ‘r’ is released from rest from the top of a wedge of mass ‘2m’ initially at rest on a smooth horizontal surface as shown. There is sufficient friction between the sphere and the wedge so that the sphere rolls on the wedge without slipping. The sphere leaves the wedge horizontally at A after descending through a vertical displacement h. Let   V1,V2  and  ω be the magnitudes of the velocity of the centre of sphere, velocity of wedge and the angular velocity of the sphere at the instant the sphere leaves the wedge. Choose the correct statements.

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a

V1=5gh6

b

V1+V2=rω

c

The ratio of total kinetic energy of the sphere and to the wedge at the instant the sphere leaves the wedges is 19:5

d

Total mechanical energy of sphere and wedge system is conserved during the process

answer is A, B, C, D.

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Detailed Solution

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Condition for rolling:
 v1+v2=Rww3v12R CLM;mv1=2mv2v2=v12 kwedge=12×2mv22=mv124 ksphere=12mv12+12×25mR2×(3v12R)2=1920mv12 Ratio=195  COE:mv124{1+195}=mgh   v1=5gh6
 
 
 
 
 
 
 

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